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I understand this notation is now a differential operator and this is the limit of a quotient, but Leibniz regarded $\frac {dy}{dx}$ as a quotient. In Leibniz's theory where $\frac {dy}{dx}$ is a quotient, do the terms in chain rule cancel out?

For instance below, there are two instances of $du$. $$ y = f(u), u = u_1 = u_2 = g(x) $$ $$ \frac {dy}{du_2} \Big|_{g(x)} \cdot \frac {du_1}{dx} \Big|_{x} = (f \circ g)'(x) = (f' \circ g)(x) \cdot g'(x) $$

It's my understanding that: $$(du_1 = g(x+h) - g(x)) \ne (du_2 = (g(x) +h) - g(x))$$ becomes the following under the infinitesimal theory Leibniz used: $$\lim_{h \rightarrow 0} {(du_1 = g(x+h) - g(x)) = (du_2 = (g(x) +h) - g(x))}$$ $$du_1 =du_2$$

and the above terms $du_1, du_2$ cancel out, leaving $\frac {dy}{dx}$.

If these terms do not cancel and are unequal, why is $du_1 = du_2 = du$ used in the definition of the chain rule in modern theory? Are there advantages to viewing things like this from infinitesimal theory like Leibniz?

Nick
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  • what are u_1 and u_2? – tryst with freedom Feb 27 '21 at 20:37
  • How many times did you already ask a question regarding Leibniz notation :)? – vitamin d Feb 27 '21 at 20:39
  • This is a different question – Nick Feb 27 '21 at 20:40
  • @Buraian added them in – Nick Feb 27 '21 at 20:41
  • For differentiable functions, an infinitesimal change of $h$ gives that $g(x+h)$ is in fact infinitesimally close to $g(x)$, and hence is “equal” to $g(x)+h$. So if you are playing with actual infinitesimals, which is what Leibniz was doing, then you do have equality. It only fails to become an equality when you no longer work with infinitesimals, at which point the notation no longer represents an actual ratio and so there is no talk of “cancellation” anyway. – Arturo Magidin Feb 28 '21 at 01:08
  • @ArturoMagidin so there is no equality using current theory due to $\frac {dy}{dx}$ being the limit of a quotient, but with Leibniz's theory of infinitesimals $\frac {dy}{dx}$ was a quotient, the equality above $du_1 = du_2$ holds and therefore $du$ cancels out. It's my understanding things can be setup like Leibniz's theory with non-standard analysis. I was curious about that for awhile! – Nick Feb 28 '21 at 01:59
  • @Nick: One has to be careful: it is not immediate the $g(x+h) = g(x)+h$ holds when $h$ is an infinitesimal: you have to actually prove it (it’s essentially the fact that $g$ is continuous because it is differentiable). Only after you prove it do you get to “cancel” them in non-standard analysis. – Arturo Magidin Feb 28 '21 at 02:00
  • @ArturoMagidin do you know if viewing the theory as Leibniz did provides any helpful insight into other items, similar to the chain rule cancelling out? I have found little resource on this. – Nick Feb 28 '21 at 02:00
  • @Nick: “Vieweing the theory as Leibniz did” leads to logical contradictions. It takes some care to clean it up properly. The “resources” are non-standard analysis. Leibniz notation is useful because it is intuitive, but you still have prove the things it “suggests”. I already mentioned, in that other post, some examples, like the formula for the derivative of the inverse. You don’t get a free lunch out of Leibniz notation: you just push the issues to a different place. – Arturo Magidin Feb 28 '21 at 02:02
  • @ArturoMagidin per above, if considered in the theory of infinitesimals is $\frac {dy} {dx} = \frac {f(g(x+h) - g(x))}{g(x+h) - g(x)}$? – Nick Apr 03 '21 at 02:29
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    @Nick. Of course not. Why would you get that? You get $\frac{f(g(x+h))-f(g(x))}{(x+h)-h}$, If $g(x+h)=g(x)+h$ (using continuity), then you can rewrite as $\frac{f(g(x)+h)-f(g(x))}{h} = \frac{f(g(x)+h)-g(x)}{g(x)+h-g(x)}\cdot\frac{g(x+h)-g(x)}{h}$. – Arturo Magidin Apr 03 '21 at 02:44
  • @ArturoMagidin did you mean : $\frac {f(g(x)+h) - f(g(x))}{g(x)+h-g(x)} \cdot \frac{g(x+h) -g(x)}{h} = \frac {f(g(x) +h) - f(g(x))}{h}$? I think the numerator of first term after the equality in your comment is incorrect – Nick Apr 03 '21 at 02:59
  • @Nick: Yes, there is an $f$ missing in the subtrahend of that numerator. – Arturo Magidin Apr 03 '21 at 03:05

1 Answers1

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There is no equality using the widely used theory of limits because $\frac {dy}{dx}$ is the limit of a quotient, however with Leibniz's theory of infinitesimals $\frac {dy}{dx}$ is a quotient. Leibniz's infinitesimal theory has logical contradictions and to resolve them the theory of non-standard analysis on the hyperreal numebers can be used. The hyperreals $\mathbb{R}*$ are an extension to the reals with infinitesimals $\mathbb{R} \subseteq \mathbb{R}^*$. The derivative and integral of calculus are expressed with the standard part function which is $s : \mathbb{R}^* \rightarrow \mathbb{R}$.

With non-standard analysis the above equality $du_1 = du_2 = g(x+h) = g(x) + h$ holds if it is proven. For instance $g(x)$ is continuous because it is differentiable. When proven $du$ cancels out. When terms are cancelled out due to equality the following is the result:

$$y = f(x), u = g(x)$$ $$ \frac {dy}{du} \frac {du}{dx} = \frac {f(g(x+h)) - f(g(x)))}{g(x+h)-g(x)} \frac {g(x+h) -g(x)}{h} = \frac {f(g(x+h) -g(x))}{h} $$

due to continuity, $g(x) + h - g(x) = g(x+h) - g(x)$ then the result is: $$ \frac {f(g(x) + h) - f(g(x))} {h} $$

This appears to agree with the result using limits: $$ f'(g(x)) = \lim_{h\rightarrow \infty}{\frac {f(g(x) + h) - f(g(x))} {h}} $$

Nick
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